'''
Given a rows x cols screen and a sentence represented by a list of non-empty words, find how many times the given sentence can be fitted on the screen.

Note:

A word cannot be split into two lines.
The order of words in the sentence must remain unchanged.
Two consecutive words in a line must be separated by a single space.
Total words in the sentence won't exceed 100.
Length of each word is greater than 0 and won't exceed 10.
1 ≤ rows, cols ≤ 20,000.
Example 1:

Input:
rows = 2, cols = 8, sentence = ["hello", "world"]

Output: 
1

Explanation:
hello---
world---

The character '-' signifies an empty space on the screen.
Example 2:

Input:
rows = 3, cols = 6, sentence = ["a", "bcd", "e"]

Output: 
2

Explanation:
a-bcd- 
e-a---
bcd-e-

The character '-' signifies an empty space on the screen.
Example 3:

Input:
rows = 4, cols = 5, sentence = ["I", "had", "apple", "pie"]

Output: 
1

Explanation:
I-had
apple
pie-I
had--

The character '-' signifies an empty space on the screen.
'''

class Solution(object):
	def wordsTyping(self, sentences, rows, cols):
		""""
		:sentences List<String>
		:rows int
		:cols int
		"""
		sentence = '-'.join(sentences)
		sentence += '-'

		index_in_sentence = 0
		for row in range(rows):
			index_in_sentence += cols
			if sentence[(index_in_sentence%len(sentence))] == '-':
				index_in_sentence += 1
			else:
				while index_in_sentence > 0 and sentence[((index_in_sentence - 1)%len(sentence))] != '-':
					index_in_sentence -= 1

		return index_in_sentence/len(sentence)

solution = Solution()
row, col = 3, 6
sentences = ["a", "bcd", "e"]
print solution.wordsTyping(sentences=sentences, rows=row, cols=col)
